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Three particles, each of the mass m are ...

Three particles, each of the mass `m` are situated at the vertices of an equilateral triangle of side `a`. The only forces acting on the particles are their mutual gravitational forces. It is desired that each particle moves in a circle while maintaining the original mutual separation `a`. Find the initial velocity that should be given to each particle and also the time period of the circular motion. `(F=(Gm_(1)m_(2))/(r^(2)))`

Text Solution

Verified by Experts

The correct Answer is:
A
`F_(A) = F _(AB) +F_(AC)`
`= 2[(Gm^(2))/(a^(2))]cos30^(@) = sqrt 3[(Gm^(2))/(a^(2))]`
`r = (a)/(sqrt3)`,
`(m upsilon^(2))/(r) = F` or `(sqrt3m upsilon ^(2))/(a) = (sqrt3Gm^(2))/(a)`
`upsilon = sqrt ((Gm)/(a))`.
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Knowledge Check

  • Three particles each of mass m are kept at vertices of an equilateral triangle of side L. The gravitational field at centre due to these particle is

    A
    zero
    B
    `(3GM)/(L^(2))`
    C
    `(9 GM)/(L^(2))`
    D
    `(12 GM)/(sqrt(3)L^(2))`

  • Three particles each of mass m are kept at the vertices of an euilateral triangle of side L . The gravitational field at the centre due to these particle is

    A
    zero
    B
    `(3GM)/(L^(2))`
    C
    `(9GM)/(L^(2))`
    D
    `(2GM)/(L^(2))`

  • Three particles of equal mass 'm' are situated at the vertices of an equilateral triangle of side L . The work done in increasing the side of the triangle to 2L is

    A
    `(2G^(2)m)/(2L)`
    B
    `(Gm^(2))/(2L)`
    C
    `(3Gm^(2))/2L`
    D
    `(3Gm^(2))/(L)`